Freeman Dyson proposed that the power needs of an advanced civilisation would eventually require the entire energy output of a star to be collected, so that the star would end up surrounded by a dense network of satellites extracting power from the radiation. In science fiction this idea has often mutated into a solid shell that completely surrounds the star. My question is, if it were possible to build such a shall, what effect would it have on the dynamics of the star inside?

I suspect the effect would be catastrophic, but I'd like more details. Here's my reasoning: Wikipedia implies that the temperature gradient between the core and outer layers of a star plays an important role in its stability. The shell around the star would reflect or re-emit a lot of the star's radiation. (I'm assuming the shell isn't composed of perfect solar collectors that would simply absorb all the radiation.) This reflected radiation would reduce the loss of heat from the outer layers of the star. This would reduce or even eliminate the temperature gradient, which I guess would cause the star to expand. My question is whether this is correct, and if so whether it would be enough to disrupt the process of fusion in the star's core. Or would there be some other, less obvious effect on the star's dynamics?

A closely related question is, would stars be stable in a static or contracting universe? In this case all of space would be filled with the radiation emitted from other stars, and I'd be interested to know what effect this would have on stellar dynamics.


Main sequence stars of roughly one solar mass are--in a lot of ways--really, really simple objects.

They have non-convecting cores and non-convecting envelopes and may have a convecting layer in the middle.

About the only things that a uniform change in the external temperature could do is tweak the location and existence of the convecting band and marginally increase the overall temperature.

That might have a measurable effect on the star's lifecycle (how fast it burns, how much of the outer hydrogen makes in down to the core to fuse before the switch to helium burning), but to first order I would expect the changes to be mediated by $T_{ext}/T_{star}$, and so to be small. (BTW--I'm suspect, but can't prove, that the $T_{star}$ you want to use depends on what you're looking at (i.e. the middle layers for convection band tweaks, but the core temperature for fusion rate tweaks).

  • $\begingroup$ Thanks for the answer. I've accepted it, but I'm still interested in what happens if the constant outside temperature is really, really high. In the case of a static universe (or a perfectly thermally insulating Dyson sphere) the radiated energy would have nowhere to go, and consequently the vacuum temperature would reach the stars' surface temperature. At that point the stars can no longer keep their outsides cool by radiating, so it could increase still further, up to the ~$10^7$ K core temperature and beyond... $\endgroup$ – Nathaniel Feb 6 '12 at 15:08
  • $\begingroup$ The other thing I guess I'm kind of curious about now is, if you wanted to destroy a star (i.e. stop the fusion process) while expending a minimal amount of energy, how would you do it? I guess that's a bit frivolous to make into a proper question though. $\endgroup$ – Nathaniel Feb 6 '12 at 15:11
  • 1
    $\begingroup$ @Nathaniel: What happens as it gets hotter is much like what happens in the core of a bigger star...it burns faster (and by faster we mean the rate grows exponentially with the temperature), switches to helium burning much sooner goes red giant with enough force to lose a lot of mass and so on. That said, destroying a star is really hard without some kind of science-fictional technology. The Charles Stross book Iron Sunrise posits one of way doing it that is pretty rigorous aside from the not-related-to-any-physics-we-know triggering mechanism. $\endgroup$ – dmckee --- ex-moderator kitten Feb 6 '12 at 18:55
  • $\begingroup$ The outer envelope of a sun-like star is convective. $\endgroup$ – Rob Jeffries Feb 15 '15 at 13:25
  • $\begingroup$ My apologies for the unaccept. It's because, thinking about this again years later, it doesn't seem right. The core of the Sun is $1.5\times 10^7$ K. If the Sun was in an adiabatic container I would expect it to equilibrate at or above that temperature. If it did so, the RMS velocity of hydrogen ions at the surface would be well above escape velocity. I guess this means the star would expand to fill the container. Perhaps fusion would slow or something before that happened, but in any case the answer of "not much really" doesn't seem correct. $\endgroup$ – Nathaniel Nov 12 '16 at 14:38

Interesting question - I don't know much about stellar dynamics but my guess is it will have very little effect.

My reasoning: right now the Sun is receiving thermal radiation from a heat bath which is dominated by the cosmic microwave background, at a temperature of about 2K. If the Dyson sphere were to be built, presumably it would be designed to operate at a temperature comfortable for human life / technology, around 300K.

So the Sun's background will rise by several hundred degrees - but this should be negligible when compared to it's own enormous temperature, so there should be little or no changes.

  • $\begingroup$ The shell temperature will be dependant on its radius. The energy radiated by the shell will be equal to the energy radiated by the star, so the increased surface area of the shell versus the star will control the temperature required for equilibrium. $\endgroup$ – Colin K Feb 5 '12 at 16:43
  • $\begingroup$ @ColinK In the case of a thermodynamically perfect system, the equilibrium temperature would be controlled by the outer radius of the shell. The other limit is controlled by the inner radius. A "real" Dyson sphere probably falls somewhere between and presumably is close to the perfect that worst case. $\endgroup$ – dmckee --- ex-moderator kitten Feb 5 '12 at 17:01
  • $\begingroup$ @dmckee: Hmm, I was thinking of it as having no thickness, but you're right. $\endgroup$ – Colin K Feb 5 '12 at 18:09
  • 1
    $\begingroup$ The radius of the sphere would determine the temperature of the outside of the sphere, assuming the star's output remains constant. The temperature inside the sphere can be hotter if the sphere is not made of a perfect thermal conductor. I'm most interested in the case where it's a perfect insulator, i.e. it reflects all of the star's radiation back at it. Naturally this wouldn't result in a habitable environment on the inside - this is just a thought experiment to see whether the "cold radiation bath" is important for the stability of a star. $\endgroup$ – Nathaniel Feb 5 '12 at 18:48
  • $\begingroup$ Not clear why you think the temperature difference matters. The heat is not transferred by conduction. It is the reflected luminosity that matters, which is to first order independent of how close the Dyson sphere is. $\endgroup$ – Rob Jeffries Feb 15 '15 at 13:38

I think this question is equivalent to saying what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much flux can actually escape from the photosphere

The global effects, depend on the structure of a star and differ for one that is fully convective, or one like the Sun that has a radiative interior and a relatively thin convective envelope on top. The phenomenon could be treated in a similar way to the effects of large starspots. The canonical paper on this is by Spruit & Weiss (1986). They show that the effects have a short term character and then a long term nature. The division point is the thermal timescale of the convective envelope, which is of order $10^{5}$ years for the Sun.

On short timescales the nuclear luminosity of the Sun is unchanged, the stellar structure remains the same as does the surface temperature. As only a fraction of the flux from the the Sun gets into space, the net luminosity at infinity will be decreased. However things change if you leave the Dyson sphere in place for longer.

On longer timescales, in a star like the Sun, the luminosity will tend to stay the same because the nuclear burning core is unaffected by what is going on in the thin convective envelope. However if a large fraction of the luminosity is being reflected back then to lose the same luminosity it turns out that the radius increases and the photosphere gets a little hotter. In this case, the radius squared times the photospheric temperature will increase to make sure that the luminosity observed beyond the Dyson sphere stays the same - i.e. by $R^2T^4(1 - \beta) = R_{\odot}^2 T_{\odot}^4$, where $\beta$ is the fraction of the solar luminosity reflected by the sphere.

The calculations of Spruit et al. (1986) indicate that for $\beta=0.1$ the surface temperature increases by just 1.4% whilst the radius increases by 2%. Thus $R^2 T^4$ is increased by a factor 1.09. This is not quite $(1-\beta)^{-1}$ because the core temperature and luminosity do drop slightly in response to the increased radius.

It is probably not appropriate to quantitatively extrapolate the Spruit treatment for very large values of $\beta$, but why would you build a Dyson sphere that was highly reflective? However, qualitatively I think the star would expand massively and I guess eventually engulf the Dyson sphere.

The above discussion is true for the Sun because it has a very thin convection zone and the conditions in the core are not very affected by conditions at the surface. As the convection zone thickens (for example in a main sequence star of lower mass), the response is different. The increase in radius becomes more pronounced; to maintain hydrostatic equilibrium the core temperature decreases and hence so does the nuclear energy generation. The luminosity of the star falls and the surface temperature stays roughly the same.


Building it as a solid shell is a really terrible idea: it would just plummet into the star due to slight imbalances in gravity. - this is wrong, see comments.

In terms of stellar dynamics, I don't think it would have any major impact on the star. There would still be a temperature gradient as you can't really heat up a vacuum. There maybe some radiation pressure due to the reflected photons but this will be pretty insignificant. The star might eventually expand but I think this will be well beyond its main sequence phase anyway.

  • 2
    $\begingroup$ Er...a solid spherical shell has neutral stability in a $1/r^2$ field (i.e. both gravitation and the radiation), so your first sentence is not really correct. It would require proportionally less stabilization than a ringworld. $\endgroup$ – dmckee --- ex-moderator kitten Feb 5 '12 at 19:43
  • $\begingroup$ If you think about it, the shell's center of mass is directly in the middle. If that is even slightly misaligned with the CoM of the star, one side of the sphere will be attracted to the star more than the other and it will move towards that direction. As the side gets closer, it is attracted even more, causing it to crash into the star. It is explained on wikipedia better than I can: en.wikipedia.org/wiki/Dyson_sphere#Dyson_shell $\endgroup$ – Durand Feb 5 '12 at 21:03
  • $\begingroup$ Durand: No, the wikipedia says that uncorrected drifts are a problem, and it is right because the sphere is not stable, but it is also not un-stable. There is no net gravitation force between the star and the sun for any relative position with the sun inside. Yes, one side is closer, but there is more stuff on the away side and it is just enough to balance out. Really. This is a well known problem. Follow the link in the wiki article so Shell theorem. By contrast a Niven ring is unstable in it's own plane (but stable out of plane). $\endgroup$ – dmckee --- ex-moderator kitten Feb 5 '12 at 21:07
  • $\begingroup$ Mm, ok. I see what you're saying. I always end up confusing myself when I think about problems like this! $\endgroup$ – Durand Feb 5 '12 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.